DBE: General Material Balance

For our balance, we will choose an arbitrary fixed (in space) control volume. Thus, it is an open system (i.e., one through which mass may flow). Accounting for mass within that volume (and transfered with surroundings) we get:

rate of accumulation of mass in CV = net rate of mass transport into CV + net rate of mass "generation" in CV

Here the "generation" term arises due to our balance being focused on a specific type of material, so reactions can generate/consume it.

In writing this mathematically, we use ci for a balance on a molar basis and for a balance based on mass. We then get:

$\displaystyle{\frac{\partial}{\partial t}\int \int \int_{CV} \rho_i dV = - \int \int_{CS} \rho_i (\mathbf{v_i\cdot n}) dA + \int \int \int_{CV} \dot r_i dV}$

If we use the definition of the mass flux vector (of component i), ${\mathbf n_i} = \rho_i {\mathbf v_i}$, we can re-write this equation as:

$\displaystyle{\frac{\partial}{\partial t}\int \int \int_{CV} \rho_i dV = - \int \int_{CS} (\mathbf{n_i\cdot n}) dA + \int \int \int_{CV} \dot r_i dV}$

Now using the same bit of math wizardry that we used before (i.e., the Gauss divergence theorem) ...

$\displaystyle{\int \int \int_{CV} \left [ \frac{\partial \rho_i}{\partial t} + \nabla \cdot \mathbf{n_i} - \dot r_i \right ] dV = 0}$

Again, since we chose an arbitrary control volume, the integrand must be zero. So, we now have a differential material balance equation:

$\displaystyle{\frac{\partial \rho_i}{\partial t} = - \nabla \cdot \mathbf{n_i} + \dot r_i}$

We could write the same equation in terms of molar concentrations by dividing by the molecular weight of the material:

$\displaystyle{\frac{\partial c_i}{\partial t} = - \nabla \cdot \mathbf{N_i} + \dot R_i}$

Simplifying the GMB

Unlike the GTEB, the General Mass Balance (GMB) equation has only three assumptions/conditions that we can make that to simplify our problems:

OUTCOME:

Simplify the general material energy balance equation (identifying assumptions)